Consider a tetrahedron `D-ABC` with position vectors if its angular points as
A(1, 1, 1), B(1, 2, 3), C(1, 1, 2)
and centre of tetrahedron `((3)/(2), (3)/(4),2)`.
Q. If N be the foot of the perpendicular from point D on the plane face ABC then the position vector of N are :

Options

A. (-1, 1, 2)
B. (1, -1, 2)
C. (1, 1, -2)
D. (-1, -1, 2)

Answer

To find the position vector of point N, which is the foot of the perpendicular from point D to the plane defined by points A, B, and C, we can follow these steps:

### Step 1: Identify the Coordinates of Points A, B, C, and the Center of the Tetrahedron
The coordinates of the points are given as:
- A(1, 1, 1)
- B(1, 2, 3)
- C(1, 1, 2)

The center of the tetrahedron is given as:
- Center = \(\left(\frac{3}{2}, \frac{3}{4}, 2\right)\)

### Step 2: Determine the Coordinates of Point D
Let the coordinates of point D be \((x_1, y_1, z_1)\).

Using the formula for the center of a tetrahedron:
\[ \text{Center} = \frac{A + B + C + D}{4} \]
We can set up the following equations based on the coordinates of the center:

For the x-coordinate:
\[ \frac{1 + 1 + 1 + x_1}{4} = \frac{3}{2} \]
Multiplying through by 4:
\[ 3 + x_1 = 6 \implies x_1 = 3 \]

For the y-coordinate:
\[ \frac{1 + 2 + 1 + y_1}{4} = \frac{3}{4} \]
Multiplying through by 4:
\[ 4 + y_1 = 3 \implies y_1 = -1 \]

For the z-coordinate:
\[ \frac{1 + 3 + 2 + z_1}{4} = 2 \]
Multiplying through by 4:
\[ 6 + z_1 = 8 \implies z_1 = 2 \]

Thus, the coordinates of point D are \(D(3, -1, 2)\).

### Step 3: Find the Equation of Plane ABC
To find the equation of the plane formed by points A, B, and C, we can use the determinant method. The general form of the equation of a plane given three points \((x_1, y_1, z_1)\), \((x_2, y_2, z_2)\), and \((x_3, y_3, z_3)\) is:
\[ \begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0 \]

Substituting the coordinates of points A, B, and C:
\[ \begin{vmatrix} x - 1 & y - 1 & z - 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{vmatrix} = 0 \]

Expanding this determinant:
\[ (x - 1) \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} - (y - 1) \begin{vmatrix} 0 & 2 \\ 0 & 1 \end{vmatrix} + (z - 1) \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} = 0 \]
This simplifies to:
\[ (x - 1)(1) = 0 \implies x = 1 \]

### Step 4: Find the Foot of the Perpendicular N from D to Plane ABC
Since the plane ABC is defined by \(x = 1\), the foot of the perpendicular N from point D(3, -1, 2) to this plane will have the same y and z coordinates as point D, but the x-coordinate will be 1.

Thus, the coordinates of point N are:
\[ N(1, -1, 2) \]

### Final Answer
The position vector of point N is:
\[ \vec{N} = \langle 1, -1, 2 \rangle \]